Why does K\rightsquigarrow K(X) preserve the degree of field extensions?

Step 1
Let ${\displaystyle \frac{K}{k}}$ be a field extension.
Then
${\displaystyle K{\otimes }_{k}k\left(X\right)}$
is a localization of the integral domain
${\displaystyle K{\otimes }_{k}k\left[X\right]=K\left[X\right]}$
thus the natural map
${\displaystyle K{\otimes }_{k}k\left(X\right)\to K\left(X\right)}$
is injective. Now we have the following equivalence:
a) ${\displaystyle \frac{K}{k}}$ is algebraic
b) Every polynomial over K divides some polynomial over k.
c) ${\displaystyle K{\otimes }_{k}k\left(X\right)=K\left(X\right)}$
${\displaystyle a\Rightarrow b}$: Because of uniqueness of the division algorithm, it suffices to prove divisibilty in an algebraic extension. So take a splitting field of the polynomial and reduce to the case of linear factors: ${\displaystyle x-a}$. But they divide the minimal polynomial over k.
${\displaystyle b\Rightarrow c:K{\otimes }_{k}k\left(X\right)}$
is the localization of ${\displaystyle K\left[X\right]}$ at all nontrivial polynomials over k. Since we have b, we localize at all nontrivial polynomials over K, i.e. we get ${\displaystyle K\left(X\right)}$.
${\displaystyle c\Rightarrow a:}$
Take ${\displaystyle u\in K}$.
Then ${\displaystyle \frac{1}{(X-u)}=\frac{p}{q}}$
for
${\displaystyle p\in K\left[X\right]}$
${\displaystyle q\in k\left[X\right]}$
i.e. ${\displaystyle p(X-u)}$
is a polynomial over k with root u. QED
In particular, if ${\displaystyle \frac{K}{k}}$ is algebraic, we have ${\displaystyle [K\left(X\right):k\left(X\right)]=[K:k]}$

Step 1
There is also a more elementary argument explaining why
${\displaystyle v_1,\cdots ,v_n}$ spans ${\displaystyle \frac{K\left(X\right)}{k\left(X\right)}}$. To understand it one doesn't need to know separable field extensions nor integral ring extensions, only a little bit of linear algebra:
Let ${\displaystyle E={\text{span}}_{k\left(X\right)}\{v_1,\cdots ,v_n\}}$. Note that ${\displaystyle K\left[X\right]\subset E}$ implies ${\displaystyle K\left[X\right]E\subset E,1\in E}$, as ${\displaystyle v_1,\cdots ,v_n}$ is a basis of ${\displaystyle \frac{K}{k}}$
Take an arbitrary ${\displaystyle h\in K\left[X\right]}$
Then
$\cdot h:K(X)\ni a\mapsto a\cdot h\in K(X)$
is an isomorphism of linear spaces over ${\displaystyle k\left(X\right)}$ (${\displaystyle h^{-1}}$ defines inverse linear mapping ${\displaystyle \cdot h^{-1}}$). Fuethermore ${\displaystyle \cdot h\left(E\right)\subset E}$ as ${\displaystyle K\left[X\right]\subset E}$
Hence
${\displaystyle \cdot h{\mid }_E:E\to E}$
is a monomorphism of finite dimensional linear spaces over ${\displaystyle k\left(X\right)}$, so in fact an isomorphism. Therefore ${\displaystyle \cdot h^{-1}{\mid }_E:E\to E}$
In particular ${\displaystyle (\cdot h^{-1})\left(1\right)=h^{-1}\in E}$, and ${\displaystyle g{h}^{-1}\in E}$ for every ${\displaystyle g\in K\left[X\right]}$

Let $\frac{K}{k}$ be an extension and X an indeterminate. We identify $k(X){\otimes }_{k}K$ with $k(X){\otimes }_{k[X]}K[X]$, and denote this domain by A. Everything will follow from the Observation. Any element of A can be written as $\frac{1}{p(X)}\otimes P(X)$ with $p(X)$ in $k[X],p(X)\ne 0,P(X)$ in $K[X]$. Its image in $K(X)$ is $\frac{P(X)}{p(X)}$. In particular, the natural map from A to $K(X)$ is injective. Thus we can (and will) view A as the subdomain of $K(X)$ formed by the elements which can be written, in the above notation, as $\frac{P(X)}{p(X)}$. The fraction field of A, regarded as a subfield of $K(X)$, is called the compositum of $k(X)$ and K. It is denoted $k(X)K$, and coincides with $K(X)$. If a is in K and $\frac{1}{(X-a)}$ is in A, then a is algebraic over k. So, A is not a field if $\frac{K}{k}$ is transcendental. If $\frac{K}{k}$ has finite degree, the domain A, being finite dimensional over $k(X)$, is a field, and is thus equal to $K(X)$. The same conclusion holds if K is a union of finite degree extensions of k, that is, if $\frac{K}{k}$ is algebraic.